Simply Connected Open Manifold being Contained in Simply Connected Closed Manifold I am aware of the work of Siebenmann characterizing when an open manifold can imbedded in a compact manifold with boundary, but I am having trouble understanding the simpler case when the compact manifold has empty boundary and both the open manifold and compact manifold are simply connected. In this case is there any simplification to Siebenmann's requirements and is there a unique way to put on a boundary?
 A: Let $M$ be a manifold with boundary. Here's a trick you can use if $\partial M$ is connected.
The double of $M$ is defined as $D(M) = M\cup_{\partial M} M$; it is a closed manifold of the same dimension. If $X$ is an open manifold which embeds in $M$, then it embeds in $D(M)$. Using the fact that $\partial M$ is connected, it follows from the Seifert-van Kampen Theorem that if $M$ is simply connected, then so is $D(M)$.
Note, this trick doesn't work if $\partial M$ is disconnected. For example, the open disc $D^n$ embeds in $M = S^{n-1}\times [0,1]$, so it embeds in the double $D(M) = S^{n-1}\times S^1$, but this is no longer simply connected.
A: Let me answer the second part of your question, namely, about the uniqueness of the boundary. More precisely, the setup for Siebenmann's compactification is that you have an open manifold $M$ (every component is noncompact). Then Siebemnann gives you some conditions under which there exists a compact manifold necessarily with nonempty boundary $N$ such that $M$ is homeomorphic to the interior of $N$, i.e. to $N- \partial N$. In this situation one can ask if $\partial N$ is uniquely determined by $M$. The answer to this question is negative, see Danny Ruberman's example here. 
